Odpowiedź:
a)
[tex] \frac{2 \sqrt{3} }{3 \sqrt{5} } = \frac{2 \sqrt{3} \times \sqrt{5} }{3 \sqrt{5} \times \sqrt{5} } = \frac{2 \sqrt{15} }{3 \sqrt{25} } = \frac{2 \sqrt{15} }{3 \times 5} = \frac{2 \sqrt{15} }{15} [/tex]
b)
[tex] \frac{ \sqrt{4} }{ \sqrt{10} } = \frac{2 \times \sqrt{10} }{ \sqrt{10} \times \sqrt{10} } = \frac{2 \sqrt{10} }{10} = \frac{ \sqrt{10} }{5} [/tex]
c)
[tex] \frac{ \sqrt{3} \times \sqrt{6} }{ \sqrt{2} } = \frac{ \sqrt{3 \times 6} }{ \sqrt{2} } = \frac{ \sqrt{18} }{ \sqrt{2} } = \frac{ \sqrt{18} \times \sqrt{2} }{ \sqrt{2} \times \sqrt{2} } = \frac{ \sqrt{36} }{2} = \frac{6}{2} = 3[/tex]
d)
[tex] \sqrt{ \frac{1}{27} } \times \sqrt{3} - ( - \frac{1}{2} ) {}^{0} = \sqrt{ \frac{1}{27} \times 3 } - 1 = \sqrt{ \frac{3}{27} } - 1 = \sqrt{ \frac{1}{9} } - 1 = \frac{1}{3} - 1 = - \frac{2}{3} [/tex]
e)
[tex] \sqrt{6} \times \sqrt{12} = \sqrt{72} = \sqrt{36 \times 2} = 6 \sqrt{2} [/tex]
f)
[tex] \sqrt{35} \div \sqrt{28} = \sqrt{35 \div 28} = \sqrt{1.25} = \sqrt{ \frac{5}{4} } = \frac{ \sqrt{5} }{2} [/tex]
